/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 An electric dipole with dipole m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electric dipole with dipole moment \(\vec{p}\) is in a uniform external electric field \(E\). (a) Find the orientations of the dipole for which the torque on the dipole is zero. (b) Which of the orientations in part (a) is stable, and which is unstable? (Hint: Consider a small rotation away from the equilibrium position and see what happens.) (c) Show that for the stable orientation in part (b), the dipole's own electric field tends to oppose the external field.

Short Answer

Expert verified
For (a), the orientations for which the torque on the dipole is zero are when \(\theta = 0\) and \(\theta = 180^\circ\). For (b), the stable equilibrium is when \(\theta = 0\), while the unstable equilibrium is when \(\theta = 180^\circ\). For (c), at the stable position, the electric field of the dipole tends to oppose the external electric field.

Step by step solution

01

Determine the Orientations for Zero Torque

The torque \(\tau\) on a dipole in an external electric field is given by the equation \(\tau = pE \sin \theta\), where \(\theta\) is the angle between the dipole moment vector and the electric field vector. Torque is zero when \(\sin \theta = 0\), which happens when \(\theta = 0\) (dipole aligns with the electric field direction) or \(\theta = 180^\circ\) (dipole aligns in the opposite direction to the electric field).
02

Determine the Stability and Instability of the Orientations

For the stability determination, consider a small rotation away from the equilibrium positions \(\theta = 0\) and \(\theta = 180^\circ\). Now, after rotating, for \(\theta = 0^\circ\), the torque will tend to rotate the dipole back to its original position, making it a stable equilibrium. On the other hand, after rotating for \(\theta = 180^\circ\), the torque will tend to rotate in the same direction of rotation , making it an unstable equilibrium.
03

Show the Opposition of the Dipole’s Electric Field to the External Field

Consider the stable orientation, which is when \(\theta = 0^\circ\). The electric field \(E'\) produced by the dipole at large distances is opposite to the orientation of the dipole. So, for this stable condition, the external electric field is aligned with the dipole, and thus the electric field produced by the dipole itself points in the opposite direction. Thus, the field of the dipole opposes the external electric field.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque on Electric Dipole
When an electric dipole with a dipole moment \( \vec{p} \) is placed in a uniform electric field \( E \) it experiences a torque \( \tau \). This torque is calculated by the cross-product of the dipole moment and the electric field, leading to the formula \( \tau = pE \sin(\theta) \) where \( \theta \) is the angle between the dipole moment and the electric field vectors.

At the angles \( \theta = 0^\circ \) and \( \theta = 180^\circ \) the sine function is zero, hence the torque is zero, indicating equilibrium positions. Understanding the relationship between torque and the dipole's orientation is crucial for applications ranging from molecular dynamics to the functioning of electronic devices.
Electric Dipole Stability
The stability of an electric dipole in an external field can be analyzed by perturbing it slightly from its equilibrium position. If the dipole at \( \theta = 0^\circ \) is displaced, the resulting torque will act to restore it to its initial position, akin to a marble resting at the bottom of a bowl—illustrating a stable equilibrium.

Conversely, if the dipole at \( \theta = 180^\circ \) is perturbed, the produced torque will further displace it from equilibrium, resembling a marble on top of a hill—this is known as an unstable equilibrium. The behavior under small displacements is often examined in physics to ascertain stability in systems ranging from electric dipoles to mechanical constructions.
Dipole's Electric Field Opposition
In a stable configuration, where the dipole moment aligns with the electric field (\( \theta = 0^\circ \) ), any external influence is naturally opposed by the dipole's response. The electric field \( E' \) generated by the dipole opposes the applied uniform electric field \( E \) because they are in opposite directions.

This opposition is comparable to a feedback mechanism where the system acts to minimize the effect of a disturbance, which in this case is the external electric field. Recognizing this opposing nature helps in understanding phenomena such as polarization in materials and the fundamental principles of electromagnetism. This knowledge is not just theoretical but is also applied in designing sensors and other electronic components.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A proton is traveling horizontally to the right at \(4.50 \times 10^{6} \mathrm{~m} / \mathrm{s}\). (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of \(3.20 \mathrm{~cm}\). (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

A very long line of charge with charge per unit length \(+8.00 \mu \mathrm{C} / \mathrm{m}\) is on the \(x\) -axis and its midpoint is at \(x=0 .\) A second very long line of charge with charge per length \(-4.00 \mu \mathrm{C} / \mathrm{m}\) is parallel to the \(x\) -axis at \(y=10.0 \mathrm{~cm}\) and its midpoint is also at \(x=0 .\) At what point on the \(y\) -axis is the resultant electric field of the two lines of charge equal to zero?

A straight, nonconducting plastic wire \(8.50 \mathrm{~cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point \(6.00 \mathrm{~cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point \(6.00 \mathrm{~cm}\) directly above its center.

Two tiny spheres of mass \(6.80 \mathrm{mg}\) carry charges of equal magnitude, \(72.0 \mathrm{nC}\), but opposite sign. They are tied to the same ceiling hook by light strings of length \(0.530 \mathrm{~m}\). When a horizontal uniform electric field \(E\) that is directed to the left is turned on, the spheres hang at rest with the angle \(\theta\) between the strings equal to \(58.0^{\circ}\) (Fig. P21.74). (a) Which ball (the one on the right or the one on the left) has positive charge? (b) What is the magnitude \(E\) of the field?

A point charge is placed at each corner of a square with side length a. All charges have magnitude \(q\). Two of the charges are positive and two are negative (Fig. E21.38). What is the direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of \(q\) and \(a\) ?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.